Service for Solving Linear Programming Problems

and other interesting typical problems
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Example №1. Solving of a System of Linear Equations by the Gauss Elimination (One Solution)

This solution has been made using the calculator presented on the site.
Please note that the coefficients will disappear which located in the "red" positions.
Знак системы4x1+2x2- 3 x3 = - 3
5x1+3x2- 5 x3 = - 8
4x1+x2+5x3 = 22
The equation 1 multiplied by -1 is added to the equation 2.   more info
( 5 x1 + 4 x1 * ( -1) )
+ ( 3 x2 + 2 x2 * ( -1) )
+ ( -5 x3 + ( -3 x3) * ( -1) )
= -8 + ( -3) * ( -1)
This transformation will allow us to count without fractions for some time.
Знак системы4x1+2x2- 3 x3 = - 3
x1 +x2- 2 x3 = - 5
4x1+x2+5x3 = 22
The equation 2 and equation 1 are reversed.
Знак системыx1 +x2- 2 x3 = - 5
4x1+2x2- 3 x3 = - 3
4x1+x2+5x3 = 22
The equation 1 multiplied by -4 is added to the equation 2.   more info
( 4 x1 + x1 * ( -4) )
+ ( 2 x2 + x2 * ( -4) )
+ ( -3 x3 + ( -2 x3) * ( -4) )
= -3 + ( -5) * ( -4)
The "red" coefficient is zero.
Знак системыx1 +x2- 2 x3 = - 5
- 2 x2+5x3 = 17
4x1+x2+5x3 = 22
The equation 1 multiplied by -4 is added to the equation 3.   more info
( 4 x1 + x1 * ( -4) )
+ ( x2 + x2 * ( -4) )
+ ( 5 x3 + ( -2 x3) * ( -4) )
= 22 + ( -5) * ( -4)
The "red" coefficient is zero.
Знак системыx1 +x2- 2 x3 = - 5
- 2 x2+5x3 = 17
- 3 x2+13x3 = 42
The equation 3 multiplied by -1 is added to the equation 2.   more info
( -2 x2 + ( -3 x2) * ( -1) )
+ ( 5 x3 + 13 x3 * ( -1) )
= 17 + 42 * ( -1)
This transformation will allow us to count without fractions for some time.
Знак системыx1 +x2- 2 x3 = - 5
x2 - 8 x3 = - 25
- 3 x2+13x3 = 42
The equation 2 multiplied by 3 is added to the equation 3.   more info
( -3 x2 + x2 * 3 )
+ ( 13 x3 + ( -8 x3) * 3 )
= 42 + ( -25) * 3
The "red" coefficient is zero.
Знак системыx1 +x2- 2 x3 = - 5
x2 - 8 x3 = - 25
- 11 x3 = - 33
We will find the variable x3 from equation 3 of the system.
- 11 x3 = - 33
x3 = 3
We will find the variable x2 from equation 2 of the system.
x2 - 8 x3 = - 25
x2 = - 25 + 8 x3
x2 = - 25 + 8 * ( 3 )
x2 = - 1
We will find the variable x1 from equation 1 of the system.
x1 + x2 - 2 x3 = - 5
x1 = - 5 - x2 + 2 x3
x1 = - 5 - ( - 1 ) + 2 * ( 3 )
x1 = 2
Result:
x1 = 2
x2 = - 1
x3 = 3








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